Name______________________________________
EMPIRICAL FORMULA WORKSHEET
Problems that involve percentages, mass or moles of elements: Follow the poem. Show your work. Fill in the
chart. Sometimes, the multiplier is 1. After awhile, you will not need the chart to help you with the problems.
One last reminder: the empirical formula and the molecular formula are related by a whole number ratio
1. A compound is 24.7% Calcium, 1.2% Hydrogen, 14.8% Carbon, and 59.3% Oxygen. Write the empirical
formula and name the compound.
%®mass
mass®moles Divide by small Multiply until whole Final ratio
Element Mass
Molar mass moles
Mole ratio
C
12.01 g/mol
H
O
Ca
Show your work for the calculation of empirical formula here
Empirical formula and combustion analysis worksheet
Page 1 of 8
9/24/17
Name______________________________________
2.
Excessive physical activity, lactic acid molecular mass 90.08 g per mole, forms in muscle tissues and is
responsible for muscle soreness. Elemental analysis shows that this compound has 40.0% carbon 6.71%
hydrogen and 53.3% oxygen. Determine the empirical formula of lactic acid. Determine the molecular formula.
Divide by
Multiply until
Final
%®mass
mass®moles
empirical
small
whole
ratio
Element
Mass
Molar mass
moles
Mole ratio
C
12.01g/mol
H
O
Show your work for the calculation of empirical formula here
Show your work for the calculation of the molecular formula here
Empirical formula and combustion analysis worksheet
Page 2 of 8
9/24/17
Name______________________________________
3.
A compound is 21.20% Nitrogen, 6.06% Hydrogen, 24.30% Sulfur, and 48.45% Oxygen. Write the empirical
formula for the compound.
Divide by
Multiply until
Final
%®mass
mass®moles
empirical
small
whole
ratio
Element
Mass
Molar mass
moles
Mole ratio
N
H
S
O
Show your work for the calculation of empirical formula here
Empirical formula and combustion analysis worksheet
Page 3 of 8
9/24/17
Name______________________________________
4.
Nicotine is an addictive compound found in tobacco leaves. Elemental analysis of nicotine gives the following
data: 74.0 % C, 8.65 % H, 17.35 % N. What is the empirical formula of nicotine? The molar mass of nicotine
is less than 170 g/mol. What is the molecular formula?
Divide by
Multiply until
Final
%®mass
mass®moles
empirical
small
whole
ratio
Element
Mass
Molar mass
moles
Mole ratio
C
H
N
Show your work for the calculation of empirical formula here
Show your work for the calculation of the molecular formula here
Empirical formula and combustion analysis worksheet
Page 4 of 8
9/24/17
Name______________________________________
5.
A sample of an unknown compound contains 0.21 moles of zinc, 0.14 moles of phosphorus, and 0.56 moles of
oxygen. What is the empirical formula?
Divide by
Multiply until Final
%®mass
mass®moles
empirical
small
whole
ratio
Element
Mass
Molar mass
moles
Mole ratio
Show your work for the calculation of empirical formula here
Empirical formula and combustion analysis worksheet
Page 5 of 8
9/24/17
Name______________________________________
Combustion problems: combustion analysis problems are more challenging, but with a little practice and
organization of data, you will find they are similar to the percentage problems.
Some helpful rules:
a. You are working backwards from CO2 and H2O.
b. Convert grams of CO2 to grams and moles of C
c. Convert grams of H2O to grams and moles of H (not H2)
d. Mass of the C and mass of the H should add to the mass of the sample. If it does not, the mass that is
missing is oxygen (again, O not O2)
e. Convert grams of oxygen to moles of oxygen.
f. Now divide by small and multiple until whole.
g. Sometimes the element left over is not oxygen, so read the question carefully!
h. Use the chart!
i. And no, you can’t use stoichiometry. You don’t have a balanced equation.
I have started most of the problem below using the rules. Let’s finish it. You need to fill in the chart with the calculated
data. Then find the ratios. Once you find the ratios, you can find the empirical and the molecular formula.
1. Vitamin C, M= 176.12 g/mol, is a compound of composed of carbon, hydrogen, and oxygen. Vitamin C is
found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is burned in a
combustion apparatus 1.50 g of CO2 and 0.410 g of H2O are produced. What is the empirical formula of
Vitamin C? What is the molecular formula of Vitamin C?
Mass
Molar mass
moles
Mole ratio
Multiply until
Final
empirical
whole
ratio
CO2
1.50 g
44.01g/mol
H 2O
0.410 g
18.02 g/mol
C
0.4093 g
12.01g/mol
H
1.008 g/mol
O
16.00 g/mol
The answers are in bold
1.50 g
1 mol CO2
CO2
44.01 g CO2
0.410gH2O
1 mol C
0.03408
0.03408
mol C
0.03408 mol C
1 mol CO2
1 mol H2O
2 mol H
18.02g H2O
1 mol H2O
1.000g CHO — (0.0459 g H + 0.4093g C) =
12.01g C
0.4093g C
1 mol C
0.0455
mol H
0.0455 mol H
1.008g H
0.0459g H
1 mol H
0.5448g O
0.5448 g O
1 mol O
0.03405 mol O
16.00g O
Re-write set-ups, show your work, finish the vitamin C problem here
Empirical formula and combustion analysis worksheet
Page 6 of 8
9/24/17
Name______________________________________
2. Sometimes, athletes illegally use anabolic steroids to increase muscle strength. A forensic chemist analyzes
CO2
some tablets suspected of being a popular steroid. He determined that the substance in the tablets contains
only carbon, hydrogen, and oxygen in and has a molar mass of 300.14 g/mole. When a 1.200 g sample of this
study by combustion analysis, 3.516 g of CO2 and 1.007 g of H2O are collected. What is the molecular formula
for the substance?
Mass
Molar mass
moles
Mole ratio
Multiply until
Final
empirical
whole
ratio
44.01g/mol
H 2O
18.02 g/mol
C
12.01g/mol
H
1.008 g/mol
O
16.00 g/mol
g
1 mol CO2
1 mol C
44.01 g CO2
1 mol CO2
1 mol H2O
2 mol H
18.02g H2O
1 mol H2O
______mol C
_________ mol C
12.01g C
CO2
g
_________ g
C
1 mol C
Mol H
mol H
1.008g H
gH
H 2O
g CHO — (
gH+
1 mol H
g
gO
g0
1 mol O
C)
mol O
16.00g O
Empirical formula and combustion analysis worksheet
Page 7 of 8
9/24/17
Name______________________________________
3.
0.487 g sample of quinine 324g/mol is combusted and found to produce 1.321 g carbon
dioxide 0.325 g water and 0.0421 g nitrogen. Determine in empirical and molecular formula
(hint there is one element missing!)
4. After combustion with excess oxygen, a 12.501 g of a petroleum compound produced
38.196 g of carbon dioxide and 18.752 of water. A previous analysis determined that the
compound does not contain oxygen. Establish the empirical formula of the compound
5. A hydrocarbon fuel is fully combusted with 22.000 g of oxygen to yield 23.118 g of carbon
dioxide and 4.729 g of water. Find the empirical formula for the hydrocarbon.
6. 12.915 g of a biochemical substance was burned in an atmosphere of 50.123 g of oxygen.
Subsequent analysis of the gaseous result yielded 18.942 g carbon dioxide, 7.749 g of water
and 36.347 g of oxygen. Determine the empirical formula of the substance.
7. A 12.00 gram sample of a smelly compound was tested by combustion analysis. The
products were 21.41 grams of CO2, 14.59 grams of H2O, and 17.51 grams of N2O5. Further
analysis showed that oxygen was NOT present in the molecule. What is the empirical
formula of the compound?
8. When a 10.0 g sample of an unknown organic acid is subjected to combustion analysis 21.2
grams of CO2 and 3.25 g of H2O are produced. What is the empirical formula of the acid?
Answers:
3. C10H12N0, C20H24N2O2
4. C5H12 when I re-did the answer, I noticed that the ratio was off. The answer is not C3H7
5. CH, there is no oxygen in the compound and the amount of oxygen reacted is excess.
6. CH2O, oxygen is in excess. You can’t use it to find the amount of oxygen in the compound.
7. C3H10N
8. C4H3O2
Empirical formula and combustion analysis worksheet
Page 8 of 8
9/24/17